SudokuSolver Forum

Here is a Clover puzzle (combination of Asterisk and Windoku) from Peter Sudoku that I was working on a few weeks ago:



Here's the sudoku code:

.....91..2..4...7.7.5...2.9......3..37........8.1................................

Here's what JSudoku gives as the pencil marks:



However, I was unable to solve the puzzle using these pencil marks. I don't recall now what prompted me to check the pencil marks, but since I was on a 10.5-hour flight with no access to a computer, I worked out the pencil marks by hand and (successfully) redid the puzzle. These are the additional pencil marks I found:

7 in r1c4
3 in r1c5
3 in r1c9
9 in r4c1
2 in r4c9
9 in r5c7
1 and 9 in r5c8
7 in r9c3
3 in r9c5
1 and 9 in r9c8
3 in r9c9

Are these incorrect, or did I find a bug in JSudoku's pencil mark algorithm?

Hello,
JSudoku is right, there should NOT be the pencil marks you mention.

There are hidden windows in Windoku as pointed by Para at "Bank Holiday 0808 Kazaguruma Special on Mon Mar 02,2009 10:36 pm: (here http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=13&t=236&start=10)


The hidden windows mentioned are at
R1C234+R5C234+R9C234 colored brown
R1C678+R5C678+R9C678 colored blue
R234C1+R234C5+R234C9 colored purple
R678C1+R678C5+R678C9 colored grey
R1C159+R5C159+R9C159 colored white!


If we use those colors to identify the hiden windows in your case (you have to change the numbers in Para example by the numbers of your puzzle!) we have:
a) 7 in r1c4 is not possible beacuse 7 in R5C2 is already in the brown window
b) 3 in r1c5 in not possible because 3 in R5C1 is already in the white window
c) 3 in r1c9 the same as b) above
d) 9 in r4c1 is not possible because 9 in R3C9 is already in the purple window
e) 2 in r4c9 is not possible because 2 in R2C1 is already in the purple window
f) 9 in r5c7 here it is the 9 in R1C6 in the blue window that erases this pencil mark
g) 1 and 9 in r5c8 both the 9 in R1C6 and the 1 in R1C7 are in the blue window, so no other 1 or 9 in that window
h) 7 in r9c3 not possible due to 7 in R5C2 in the brown window
i) 3 in r9c5 once again the 3 in R5C1 blocks all 3 in the white window
j) 1 and 9 in r9c8 same as g) above
k) 3 in r9c9 same as b), c) and i) above

The grey window had no numbers, so it gave no problem this time, but has to be considered too when placing pencil marks.

I had TUSP calculate the pencil marks, and it came to the same result as JSudoku. TUSP does currently not support the combination of of Asterisk and Windoku, so I manually removed {257} from the 6 other (red) Asterisk cells.
Here is what I got, which is identical to JSudoku.



SOLUTION:
Using the calculated pencil marks I was able to solve the sudoku (in the second attempt).
What was not so nice is that my program crashed due to an "index out of range", when double clicking on r4c4 to set solution from Hidden Single".
Have to look at the code.

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